Poisson Distribution of sum of two random independent variables $X$, $Y$

This only holds if $X$ and $Y$ are independent, so we suppose this from now on. We have for $k \ge 0$: \begin{align*} P(X+ Y =k) &= \sum_{i = 0}^k P(X+ Y = k, X = i)\\ &= \sum_{i=0}^k P(Y = k-i , X =i)\\ &= \sum_{i=0}^k P(Y = k-i)P(X=i)\\ &= \sum_{i=0}^k e^{-\mu}\frac{\mu^{k-i}}{(k-i)!}e^{-\lambda}\frac{\lambda^i}{i!}\\ &= e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \frac{k!}{i!(k-i)!}\mu^{k-i}\lambda^i\\ &= e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \binom ki\mu^{k-i}\lambda^i\\ &= \frac{(\mu + \lambda)^k}{k!} \cdot e^{-(\mu + \lambda)} \end{align*} Hence, $X+ Y \sim \mathcal P(\mu + \lambda)$.

Another approach is to use characteristic functions. If $X\sim \mathrm{po}(\lambda)$, then the characteristic function of $X$ is (if this is unknown, just calculate it) $$ \varphi_X(t)=E[e^{itX}]=e^{\lambda(e^{it}-1)},\quad t\in\mathbb{R}. $$ Now suppose that $X$ and $Y$ are independent Poisson distributed random variables with parameters $\lambda$ and $\mu$ respectively. Then due to the independence we have that $$ \varphi_{X+Y}(t)=\varphi_X(t)\varphi_Y(t)=e^{\lambda(e^{it}-1)}e^{\mu(e^{it}-1)}=e^{(\mu+\lambda)(e^{it}-1)},\quad t\in\mathbb{R}. $$ As the characteristic function completely determines the distribution, we conclude that $X+Y\sim\mathrm{po}(\lambda+\mu)$.

You can use Probability Generating Function(P.G.F). As poisson distribution is a discrete probability distribution, P.G.F. fits better in this case.For independent X and Y random variable which follows distribution Po($\lambda$) and Po($\mu$). P.G.F of X is \begin{equation*} \begin{split} P_X[t] = E[t^X]&= \sum_{x=0}^{\infty}t^xe^{-\lambda}\frac{\lambda^x}{x!}\\ &=\sum_{x=0}^{\infty}e^{-\lambda}\frac{(\lambda t)^x}{x!}\\ &=e^{-\lambda}e^{\lambda t}\\ &=e^{-\lambda (1-t)}\\ \end{split} \end{equation*} P.G.F of Y is \begin{equation*} \begin{split} P_Y[t] = E[t^Y]&= \sum_{y=0}^{\infty}t^ye^{-\mu}\frac{\mu^y}{y!}\\ &=\sum_{y=0}^{\infty}e^{-\mu}\frac{(\mu t)^y}{y!}\\ &=e^{-\mu}e^{\mu t}\\ &=e^{-\mu (1-t)}\\ \end{split} \end{equation*}

Now think about P.G.F of U = X+Y. As X and Y are independent, \begin{equation*} \begin{split} P_U(t)=P_{X+Y}(t)=P_X(t)P_Y(t)=E[t^{X+Y}]=E[t^X t^Y]&= E[t^X]E[t^Y]\\ &= e^{-\lambda (1-t)}e^{-\mu (1-t)}\\ &= e^{-(\lambda+\mu) (1-t)}\\ \end{split} \end{equation*}

Now this is the P.G.F of $Po(\lambda + \mu)$ distribution. Therefore,we can say U=X+Y follows Po($\lambda+\mu$)